Are you accelerating while standing still on Earth's surface?

Hi Tach, przyk.

I would be very interested to read a discussion between you two on the assertion that even when standing on the Earth's surface you are effectively accelerating; as stated in the bolded/underlined part of your post below, przyk, which you made in another thread which was split/closed [for reasons not connected with this] before you had a chance to respond to my question to you there....as follows......

So, do you and Tach agree on this, przyk?....(and if not, can you and he put forth your reasoning/references to support your differing perspectives on it?)

Your discussion of this point would be much appreciated. Thanks!

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Back in a couple of days to see what transpired. Cheers.

.

 

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The ground exerts (an electromagnetic) force against your feet precluding you from falling (free) to the center of the Earth. So, you are subjected to an acceleration. GR is different from Newtonian mechanics on this subject, only free-fall in a gravitational field (or motion in the total absence of such a field) count as inertial motion. You can find this info in any GR book.

 

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The vacuum speed of light is c nearby for a free fall observer. The vacuum speed of light is also c nearby for someone standing on the earth. Though the free fall observer can be assigned a locally inertial frame, globally speaking neither observer can be given an everywhere inertial frame because of the spacetime curvature from the planet. When the spacetime curvature or Reimann tensor specifically is not zero you can not have a coordinate frame that is globally flat. So neither observer actually could be assigned a frame that is globally an inertial frame. As for the question of acceleration, though each observer has a zero coordinate acceleration according to his own choice of frame, the one standing on the surface has a nonzero invariant acceleration that all observers can agree on including himself as given by the weight reading he would have if he were on a scale divided by his mass.

 

Einstein says that you can't tell the difference if you're standing still on the ground or on a surface (in space) that's accelerating towards you, so in that sense the ground is accelerating upwards, and because of your inertia you feel weight, so you have a sense that you would fall through a hole in the surface.

 

Isn't that what he said?

 

By "he" I think you mean JR.

Actually I was pointing out that the surface accelerates towards you, which is why you feel "attracted". It's why you can't tell any difference between gravity and acceleration, or between a gravitational "push" and an inertial "pull".

 

I'm lost. When (how/why) is there acceleration without displacement?

 

Aqueous Id

The acceleration is "in balance" with the curved spacetime caused by the mass of the Earth, the electro-dynamic forces between the individual particles that make up matter changing the normal trajectory of your mass's path through spacetime toward the bottom of the gravity well. Any change to that natural trajectory is an acceleration. To electrodynamics there is no real difference in kind(you could just as easily be seen resisting the Earth's natural trajectory(which could be called the path of least energy), but our mind decides it will give deference to the larger mass, that seems more reasonable.

Grumpy

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For general relativity the acceleration four-vector is
\(A^{\lambda }=\frac{d^{2}x^{\lambda }}{d\tau ^{2}}+\Gamma ^{\lambda }_{\mu \nu }\frac{dx^{\mu }}{d\tau }\frac{dx^{\nu }}{d\tau }\)
For a free fall observer according to his own natural choice of coordinates he himself has no coordinate acceleration so
\(\frac{d^{2}x^{\lambda }}{d\tau ^{2}}=0\)
but also local to him the Christoffel symbols will vanish according to his own frame so this results in
\(A^{\lambda }=0\)
According to someone standing on the surface of a planet according to his own natural choice of coordinates he himself also has no coordinate acceleration so
\(\frac{d^{2}x'^{\lambda }}{d\tau ^{2}}=0\)
However the Chrisoffel symbols do not vanish nearby himself as evidenced by the acceleration he observes of anything he drops so his four-vector acceleration becomes
\(A'^{\lambda }=\Gamma ^{\lambda }_{ct' ct' } c^{2} \left ( \frac{dt'}{d\tau } \right )^{2} \)
\(A'^{\lambda }=\Gamma ^{\lambda }_{ct' ct' } c^{2}\)
This is not zero. The simple interpretation is that standing on the ground you know your being thrust into acceleration with respect to things nearby in free fall because you see they accelerate down with nothing pushing on them while you feel the weight of yourself from being pushed up by the ground.

 

OK thanks for that, it's out of my scope of knowledge. I grasp, superficially, the formulations that resemble standard notation for acceleration. I understand it is mapped into curved space by transformation, so I think I can catch up with you guys (at this level) with further reading.

Your statement about the non zero result being interpreted as acceleration is also puzzling to me, in the sense that I'm still looking for the displacement that acceleration refers to... although "interpretation" covers a lot of bases.

I think my difficulty is in understanding that which in the Newtonian sense is a clear distinction between a static and dynamic force, and which is (perhaps?) being merged(?) here into a force that ...doesn't distinguish between stationary and moving conditions (?) Because they are relative (?)

As you see I'm treading the margins of my knowledge. Again, thanks for going to the trouble to code and explain this to me.

Any other advice is welcome!

 

OK, thanks for putting it like that. This should help me think this through. So I can imagine a stream toward the gravity well, that I am resisting, so, that is sort of like displacement (i.e. relative to the stream I am moving?) Am I even close?

 

I thought I ought to say something here. The principle of equivalence says the guy standing on the surface of the Earth is like the guy accelerating in a spaceship. It doesn't say the guy falling to the ground is like the guy accelerating in a spaceship. It's maybe going a bit far to say you're not accelerating towards the ground when you've fallen off a ladder, and instead that you are accelerating when you're just standing there. But any complaint I'd raise about what pryzk said would revolve around his choice of words. Other than that I know what he means, and have no issue with it.

 

In my mind the curvature of space over time is such that the Earth's surface is accelerating upwards. That way all ambiguity is removed, even when you fall off your ladder.

 

Locally free fall frames have equivalent physics. The free fall observer is not actually accelerating. His acceleration four-vector is zero which yeilds geodesic motion. It is the spacetime extended around him that has is strange compared to flat spacetime.

 

Aqueous Id

The stream is actually the least energy path into the gravity well(formed by curved spacetime being distorted by mass). If you were freefalling you would be largely free of acceleration whether you were in orbit or falling straight down toward Earth, either is "going with the flow" of least energetic trajectories through spacetime for the condition in the different frames. The ground's resistence to this flow is acceleration, sort of like when you are holding onto a rock in a river resists the flow of water. The energy required to hover above the ground is the energy the ground supplies when you are standing on it, mostly from the electrodynamics between the atoms of matter.

 

I don't see it like that I'm afraid, RJ. The ground is staying where it is.

Noted. I gave the caveat about falling off the ladder to try to soften what the principle of equivalence says for layman readers.

It doesn't actually supply any energy though. The ground, or indeed your ladder, prevents the conversion of potential energy into kinetic energy. When you fall off the ladder the "force" of gravity doesn't supply any energy either. The total potential+kinetic energy is unchanged, which is why gravity isn't a force in the Newtonian sense.

 

Farsight

Correct, it is the curvature of spacetime caused by mass.

Correct, it is the energy of electrodynamic forces within the matter that resists your travelling the least energy trajectory to it's end. Take three different forms of matter and compare how this is done to see this, 1. ground, pretty well unforgiving and inelastic, has a huge excess of energy within it's structure to resist your freefall(IE the energy needed to stop your progress(IE accelerate you to a standstill)only leaves a small crater). 2. water, has much less rigidity or resistence, if you are dense enough it will only slow you down(after the initial energy of the splash), not stop your trajectory(some sink, as do most materials on Earth). Has a level of energy in it's electrodynamic bonds almost in balance with that required to stop your motion from rest in gravity(IE most people can float in it). 3. cotton candy, has some cohesiveness in the strands but is mostly just air, you would fall for a good long time before it provided enough energy to stop your progress. Or you could look at it from the other perspective, that you are expending your energy to stop the Earth from moving a tiny bit, to Relativity there is little difference in kind. And both are acceleration, even stasis is acceleration at one G.

Grumpy

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Ok, now is it true that a Newtonian "force" is actually an expression of energy (gravitational potential is expressed as an apparent force), which, if you have freedom to move (by falling off a ladder or out of an aeroplane) "converts" potential to kinetic energy.

You believe you are motionless on the surface because all the forces are in equilibrium. But your body isn't in equilibrium because you have internal motion, you see. If it didn't you wouldn't "feel" anything, like a completely static object. You would not be aware of your weight.

 

A spring under tension can be in equilibrium, yet has no internal motion.